Proof Clearly by considering the functions on a small disk $\{z \in {\mathbb C}: |z-a| \le r \}$ , where a is the intersection point of $L_1$ and $L_2$ , and by translating and rescaling as necessary, we may suppose that f and g lie in the disk algebra and the two lines intersect at $0$ .

We may also suppose that the functions are nonvanishing except possibly at $0$ .

If they do vanish at $0$ , then the orders of the zeros are equal, since for some N $\lim _{z \to 0} |f(z)|/|z|^N $ exists and is nonzero, and the same for g. So by dividing out any the zeros at $0$ , and restricting to a smaller disk if necessary, we may suppose that f and g are invertible functions in the disk algebra.

Now $h:=f/g$ is also in the disk algebra and it has modulus $1$ on both lines $L_1$ and $L_2$ . Thus, h maps $L_1$ and $L_2$ into the unit circle in the complex plane.

Suppose that for some $k \ge 1$ , we have

$$\begin{align*}h(z)=c_0+c_k z^k + O(z^{k+1}) \end{align*}$$

near $0$ , with $|c_0|=1$ and $c_k \ne 0$ ; then near to $0$ , it magnifies angles by a factor of k, and so if h is nonconstant, then $k \theta $ must be an integer multiple of $\pi $ .

Proof The conformality argument works with minor changes.

Proof Clearly, we may assume without loss of generality that $\theta =0$ . Then $h'(0) = \lim _{n \to \infty } h(a_n)/a_n \in {\mathbb R}$ . The map $h_1 : z \mapsto h(z)/z - h'(0)$ is also holomorphic around $0$ , and satisfies $h_1(a_n) \in {\mathbb R}$ . Then, in a same way, we get $h_1'(0) = h"(0) \in {\mathbb R}$ . Iterating this reasoning, we see that all the derivatives of h at $0$ must be real.

Proof We use the argument of the proof of Theorem 2.1 up to the point where we have that $h(a_n)$ and $h(b_n)$ lie in ${\mathbb T}$ for all n. By composing with suitable bilinear transformations, we can construct a function $\tilde h$ holomorphic near $0$ with sequences $c_n \to 0$ and $d_n \to 0$ such that the $c_n$ are real and the $d_n$ have argument $\theta $ , but with $\tilde h(c_n)$ and $\tilde h(d_n)$ both real for each n. By Lemma 2.4, we conclude that all the derivatives of $z \mapsto \tilde h(z)$ and $z \mapsto \tilde h(ze^{i\theta })$ are real. By looking at the Taylor series, we see that either $\tilde h$ is constant (and thus h is constant) or $e^{in\theta }=1$ for some $n>0$ .

Proof By dividing out by the common outer factor u given by

$$\begin{align*}u(z) = \exp \left( \frac{1}{2\pi} \int_0^{2\pi} \frac{e^{it}+z}{e^{it}-z} \log |f(e^{it})| \, dt \right), \end{align*}$$

we may suppose without loss of generality that f and g are inner functions.

Now, if there are any zeros of f or g on $r{\mathbb T}$ , they are at the same points, and finite in number. By dividing out a finite Blaschke product, we may suppose without loss of generality that f and g have no zeros on $r{\mathbb T}$ , so indeed $f/g$ is analytic on an annulus $\{z: r-\delta < |z| < r+\delta \}$ for some $\delta>0$ and unimodular on $r{\mathbb T}$ .

Indeed, $f/g$ has finitely many poles in $\{z: |z| < r+\delta \}$ so, multiplying by a function h such that $h(z/r)$ is a finite Blaschke product $B_1(z)$ , $fh/g$ is a function k such that $k(z/r)$ is also a finite Blaschke product $B_2(z)$ since it is analytic in a neighborhood of the disk $r{\mathbb T}$ .

Our conclusion is that $f/g$ is a rational function $F(z)=B_2(rz)/B_1(rz)$ in $r{\mathbb D}$ ; by analytic continuation (and removing any isolated singularities) $f=F g$ throughout the whole disk. But if F is nonconstant, then it is not unimodular on ${\mathbb T}$ as the zeros and poles are not placed at inverse points.

Proof This result uses a modification of an argument introduced in [Reference Chalendar, Oger and Partington1, Theorem 1.4]. Namely, we restrict F to a continuous map from ${\mathbb T}$ to the Riemann sphere ${\mathbb C} \cup \{\infty \}$ , which we continue to denote by F. Then $X = F^{-1}({\mathbb T})$ is a closed subset of ${\mathbb T}$ .

If X is an infinite proper subset of ${\mathbb T}$ , we claim that X contains a point that is a limit of two sequences, one in X and one in ${\mathbb T} \,{\backslash}\, X$ . This is clear if ${\mathbb T} \,{\backslash}\, X$ consists of a finite number of intervals. If there are infinitely many intervals, then their endpoints lie in X and accumulate at a point $z_0 \in X$ with the required property.

This point satisfies $|g(z_0)| = 1$ , and there exist two sequences $(u_n) \subset X$ and $(v_n) \subset {\mathbb T} \backslash X$ which tend to $z_0$ , such that $|F(u_n)| = 1$ and $|F(v_n)| \ne 1$ (by passing to a subsequence, we may avoid any poles of F).

By composing F with conformal mappings between ${\mathbb T}$ and ${\mathbb R} \cup \{\infty \}$ , we obtain a function h holomorphic around $0$ , and two sequences $(a_n), (b_n) \subset {\mathbb R}$ such that $a_n \to 0$ , $b_n \to 0$ , $h(a_n) \in {\mathbb R}$ and $h(b_n) \not \in {\mathbb R}$ . By Lemma 2.4, we have that all the derivatives of h at $0$ must be real, so $h(b_n) \in {\mathbb R}$ (since $b_n \in {\mathbb R}$ ). This contradiction shows that X cannot be an infinite proper subset of ${\mathbb T}$ .

Proof Let f and g have inner–outer factorizations $f=u_1 v_1$ , $g=u_2v_2$ , where $u_1$ and $u_2$ are inner and $v_1$ are $v_2$ outer with $v_1(0)>0$ and $v_2(0)>0$ . Since $|f|=|g|$ on ${\mathbb T}$ , we have $v_1=v_2$ . We may therefore suppose without loss of generality that f and g are inner.

Note that the function $F:=f/g$ is meromorphic in the disk and $|F(z)|=1$ for an infinite subset of ${\mathbb T}$ .

By Lemma 2.7, $|F|=1$ on ${\mathbb T}$ and so $|f|=|g|$ on $r{\mathbb T}$ . The result now follows from Theorem 2.6.

Proof We apply Theorem 2.8 to $f\circ \Psi _{\omega ,\alpha }$ and $g\circ \Psi _{\omega ,\alpha }$ , which implies that $f\circ \Psi _{\omega ,\alpha } =g\circ \Psi _{\omega ,\alpha }$ , and thus $f=g$ since $ \Psi _{\omega ,\alpha }$ is an automorphism of the open unit disk.

Proof Note that on $r{\mathbb T}$ , we have $\bar z=r^2/z$ ; so for an elementary Blaschke factor with $a \in {\mathbb D}$ , we have

$$\begin{align*}\frac{z-a}{1-\bar a z}\frac{\bar z-\bar a}{1-a\bar z}= \frac{(z-a)(r^2-\bar a z)}{(1-\bar a z)(z-r^2 a)}, \end{align*}$$

and thus the equation

$$\begin{align*}B_1(z)\overline{B_1(z)}=B_2(z) \overline {B_2(z)} \end{align*}$$

can be rewritten as $ R_1(z)=R_2(z)$ , where $R_1$ and $R_2$ are rational functions of degrees $2M$ and $2N$ , respectively. If $\alpha _1,\ldots , \alpha _M$ are the zeros of $B_1$ and $\beta _1,\dots , \beta _N$ are the zeros of $B_2$ , we get

$$\begin{align*}R_1(z)=\left(\frac{z-\alpha_1}{1-\overline{\alpha_1} z}\right) \cdots \left( \frac{z-\alpha_M}{1-\overline{\alpha_M} z}\right) \left( \frac{r^2-\overline{\alpha_1}z}{z-\alpha_1 r^2}\right) \cdots \left( \frac{r^2-\overline{\alpha_M}z}{z-\alpha_M r^2}\right) \end{align*}$$

and

$$\begin{align*}R_2(z)=\left(\frac{z-\beta_1}{1-\overline{\beta_1} z}\right) \cdots \left( \frac{z-\beta_N}{1-\overline{\beta_N} z}\right) \left( \frac{r^2-\overline{\beta_1}z}{z-\beta_1 r^2}\right) \cdots \left( \frac{r^2-\overline{\beta_N}z}{z-\beta_N r^2}\right).\end{align*}$$

Denote by $P_1(z)$ (resp. $P_2(z)$ ) the numerator of $R_1(z)$ (resp. $R_2(z)$ ) and $Q_1(z)$ (resp. $Q_2(z)$ ) the denominator of $R_1(z)$ (resp. $R_2(z)$ ). This reduces to a polynomial equation of degree at most $2N+2M-1$ , noting that the coefficient of $z^{2N+2M}$ of the polynomial $P_1Q_2$ is $(-1)^{N+M} \prod _{i=1}^{M} \overline {\alpha _i }\prod _{j=1}^N \overline {\beta _j}$ , which coincide with the coefficient of $z^{2N+2M}$ of the polynomial $P_2Q_1$ . Therefore, if $P_1Q_2-P_2 Q_1$ it has more than $2N+2M-1$ roots, then it is identically zero. That is, $|B_1|=|B_2|$ on the whole of $r{\mathbb T}$ and thus the result follows from the previous discussions.

Proof We have that $f/g$ is meromorphic on a neighborhood of $r\overline {\mathbb D}$ and $|f/g|=1$ on $r{\mathbb T}$ . By choosing a suitable finite Blaschke product $B_1$ and writing $\tilde B_1(z)=B_1(z/r)$ , we have that $\tilde B_1 f/g$ is still unimodular on $r{\mathbb T}$ and holomorphic on a neighborhood of $r\overline {\mathbb D}$ . Thus, it has the form $\tilde B_2$ , where $\tilde B_2(z)=B_2(z/r)$ for some finite Blaschke product $B_2$ .

Proof We may assume without loss of generality that $f,g$ are inner.

Now, the functions F and G defined by $F(s)=f(1+s)$ and $G(s)=g(1+s)$ satisfy $|F|=|G|$ on $ i{\mathbb R}$ ; and these functions are also in $H^1({\mathbb C}_+)$ , and holomorphic in a neighborhood of $i{\mathbb R}$ . Let us consider the inner–outer factorization $F=uh$ , say.

Clearly, the inner factor u has no zeros accumulating at a point on $i{\mathbb R}$ . Moreover, the outer factor h is continuous on the closed half-plane ${\mathbb C}_+$ (see, for example, the arguments in [Reference Nikolski6, Section 4.3.8]), which means that the inner factor u is also continuous except possibly at the discrete set consisting of the zeros of $uh$ on $i{\mathbb R}$ . On factoring out a zero at $iy_0$ , by considering $uh/b$ , where $b(s)= \dfrac {s-iy_0}{1+s}$ (an outer function) we see that u is also continuous at the zeros of $uh$ . Thus, u has no singular part except possibly an exponential factor $e^{-\alpha s}$ for some $\alpha \ge 0$ . We conclude that $u(s)=B_1(s)e^{-\alpha s}$ , where $B_1$ is a Blaschke product whose zeros do not accumulate at any point in $\overline {{\mathbb C}_+}$ . The function $B_1$ , which is a product of factors of the form $\dfrac {s-a_n}{s+ \overline {a_n}}$ , therefore has a meromorphic extension to ${\mathbb C}$ .

So suppose that $B_1$ has a zero at w. Then the Blaschke factor of f, say, $b_1$ , has a zero at $w+1$ , and hence a pole at $-\overline w-1$ . Thus $B_1$ has a pole at $ -\overline w-2$ and hence a zero at $w+2$ .

Repeating this argument, we find that $B_1$ has zeros at $w_n:=w+2n$ , $n \in {\mathbb N}$ . But these do not satisfy the Blaschke condition $\sum _{n=0}^\infty \frac {{\mathrm {Re}w}_n}{1+|w_n|^2} < \infty $ , and that is a contradiction.

So $B_1$ is a unimodular constant and we may apply a similar argument to the inner function v. We conclude that $f(s)/g(s)= c\exp (\lambda (s-1))$ , for a unimodular constant c, which, since f and g are inner, means that $|\exp (\lambda (iy-1))|=1$ a.e., and $\lambda =0$ .

Proof For parts (1)–(4), we may suppose without loss of generality, by composing with an automorphism, that $C_1$ is the circle $r{\mathbb T}$ centered at $0$ with radius r for some $0<r<1$ . For (1), we now consider $f_r$ and $g_r$ defined by $f_r(z)=f(z/r)$ and $g_r(z)=g(z/r)$ . Then, (1) follows easily from Corollary 2.9 and the comment following it.

For (2) with $C_1=r{\mathbb T}$ , we note from Theorem 2.12 that $f(z)/g(z)$ is a rational function of the form $B_2(z/r)/B_1(z/r)$ and hence meromorphic on ${\mathbb C} \cup \{\infty \}$ . With the change of variable $w=z/r$ , we have a meromorphic function $B_2(w)/B_1(w)$ that is unimodular on a smaller circle contained in ${\mathbb D}$ . By composition with an automorphism, we may supposed that the smaller circle is centered at $0$ . But now the fact that $f/g$ is a constant follows an argument similar to that used in the proof of Theorem 2.6 (specifically, that poles and zeros occur in pairs of inverse points with respect to one circle, which means that they cannot occur at inverse points with respect to the other circle).

(3) is easily derived from Corollary 2.14.

For (4), we may again suppose that $C_1=r{\mathbb T}$ and that $f(z)/g(z)=B_2(z/r)/B_1(z/r)$ . Now we take $w=z/r$ again, reducing to the case when a function of the form $B_2(w)/B_1(w)$ is of modulus 1 on a circle internally tangential to ${\mathbb T}$ . By means of a conformal mapping, we can transform this to the half-plane, producing a rational function that has absolute value 1 on the lines $i{\mathbb R}$ and $1+i{\mathbb R}$ . Finally, an argument based on the fact that zeros and poles must occur at pairs of inverse points a and $-\overline a$ with respect to $i{\mathbb R}$ as well as inverse points $ a$ and $-\overline a+2$ with respect to $1+i{\mathbb R}$ (cf. the proof of Theorem 2.13) leads to a contradiction unless $f/g$ is constant.

For (5), suppose that the two intersection points of $C_1$ and $C_2$ are a and b. By changing the variable to $w=1/(z-a)$ , we transform the circles into straight lines, which meet at $1/(b-a)$ , still at an angle $\theta $ . The result for irrational $\theta /\pi $ now follows from Theorem 2.1, applied to a small disk centered at $1/(b-a)$ .

We may similarly construct counterexamples for rational $\theta /\pi $ . Let $C_1$ and $C_2$ be circles of radius $1/3$ centered at $\pm 1/(3\sqrt {2})$ . Then they meet at $\pm a$ , where $a=i/(3\sqrt 2)$ . The angle between them is a right angle. The transformation $w= \dfrac {z+a}{z-a}$ sends the circles to two perpendicular lines, meeting at $0$ , from which examples similar to those in Example 2.2 can be constructed easily.